問題文全文(内容文)：
$\displaystyle (1)\, \begin{cases}
3\times(x+y-1)-4y=5 \\
5x-3(2x-y-3)=17
\end{cases}
$
$\displaystyle (2)\, \begin{cases}
0.06x+0.04y=16 \\
x+y=300
\end{cases}
$
$\displaystyle (3)\, \begin{cases}
0.2x-0.3y=0.7\\
\frac{1}{4}x+\frac{1}{3}y=\frac{1}{6}
\end{cases}
$
$\displaystyle (4)\,
5x-4y-15=3x+2y-11=-2
$
$\displaystyle (5)\,\begin{cases}
-6ax + 5by = 9\\
4bx + 3ay = 26
\end{cases}の解がx=2, \,y=3のとき、a,bを求めよ。
$

問題文全文(内容文)：
Let's enjoy oriental mathematics for just a little while.
You will discover a cute world from mathematics.
Everyone in the world should love and be good at mathematics.

We are introducing the entrance exam questions for Japanese high schools.
It's an important issue for you to understand the basics of mathematics.
Would you like to solve this math problem and check out our commentary?

問題文全文(内容文)：
$\begin{eqnarray}
\left\{
\begin{array}{l}
233x+267y=400 \\
267x+233y=100
\end{array}
\right.
\end{eqnarray}$

問題文全文(内容文)：
$ x+2y+3z=14 $
$ 2x+y+z=7 $
$ x+3y+z=10 $ を解け.

問題文全文(内容文)：
【レベル１】
①$5(2x-3y)=$
②$(8x-6y) \times (-\displaystyle \frac{1}{2})=$
③$(-16)(+10) \div (-4)=$
④$(4)(+6y)\div\displaystyle \frac{2}{3}$

【レベル２】
⑤$3(4x-2y)-(7x-5y)$
⑥$-4(-x+3y-2)-2(-5y+3x-1) $
⑦$\displaystyle \frac{2}{3}(6a-2b)+\div\displaystyle \frac{1}{3}(-9a+12b)$